Question

ajutor la trigonometrie

Answer 1

Explicație pas cu pas:

a) folosim formulele:

$sin²(A)+cos²(A)=1$

$sin(A+B)=sinAcosB+cosAsinB$

$sin(A−B)=sinAcosB−cosAsinB$

deci:

$sin²a - sin²b = sin²a - sin²a×sin²b - sin²b + sin²a×sin²b = sin²a(1 - sin²b) - sin²b(1 - sin²a) = sin²a×cos²b - sin²b×cos²a = (sina×cosb)² - (sinb×cosa)² = (sina×cosb + sinb×cosa)(sina×cosb - sinb×cosa) = sin(a+b) × sin(a-b)$

b) folosim formula:

$sinA+sinB=2sin \frac{A + B}{2} cos\frac{A - B}{2}$

deci:

$sinx + sin3x + sin5x = sin3x + (sin5x + sinx) = sin3x + 2sin \frac{5x+x}{2} cos \frac{5x-x}{2} = sin3x + 2sin3xcos2x = sin3x \times (1 + 2cos2x)$

Answer 2

Răspuns:

Explicație pas cu pas:

a)sin^2(a) -sin^2(b) = (sina -sinb)(sina +sinb)

(sina -sinb) = 2sin((a-b)/2)cos((a+b)/2)

(sina +sinb) = 2sin((a+b)/2)cos((a-b)/2)

2sin((a-b)/2)cos((a+b)/2)*2sin((a+b)/2)cos((a-b)/2) =

2sin((a-b)/2)cos((a-b)/2)*2sin((a+b)/2)cos((a+b)/2) =

sin(a-b)*sin(a+b)

b) sin5x +sinx = 2sin((5x+x)/2)cos((5x-x)/2) =

2sin3x*cos2x

2sin3x*cos2x +sin3x = sin3x(1+2cos2x)